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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle

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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 26 Mar 2015, 04:36
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 30 Mar 2015, 03:37
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 26 Mar 2015, 06:01
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Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Kudos for a correct solution.


B and C should have same Y co-ordinate --> \(4a-5 = 2a+6\) --> a=\(\frac{11}{2}\)

area of triangle = \(\frac{1}{2} * (4a-5) * (2a+1)\) . on substituting a=\(\frac{11}{2}\)
we get area = 102 .

answer A
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 26 Mar 2015, 07:14
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Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

Kudos for a correct solution.



1/2bh=1/2(2a+1)(2a+6)

Now 4a-5=2a+6
2a=11

Therefore,
A(0,0); B(0,17); C(12,17)

1/2*17*12=102

Answer: A
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 12 Jan 2018, 12:22
AmoyV wrote:
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

Now 4a-5=2a+6



Sorry, but I did not get to your calculations - why it is so? What I missed?
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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 19 Jan 2018, 02:13
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Aksena wrote:
AmoyV wrote:
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

Now 4a-5=2a+6



Sorry, but I did not get to your calculations - why it is so? What I missed?



Im guessing you have problems understanding why 4a-5 = 2a+6

Now, the reason they got to that equation was due to the given information that angle ABC = 90
--> This implies that triangle ABC was a right angle
--> Now A was given at the origin (0,0), while B was given to be somewhere along the Y-axis as its X-coordinates = 0
--> Hence, as ABC was a right triangle, the point C will have the same Y-coordinates as B in order to form the right triangle (i.e. to fulfill angle ABC = 90)
--> B(0,4a-5) and C(2a+1, 2a+6)
--> Therefore 4a-5 = 2a+6, solve for a and you can find the area of the triangle with the basic area formula for right triangles: base x height / 2
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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New post 21 Jan 2018, 19:33
Bunuel wrote:
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156


Since we have a right triangle with angle ABC = 90, we see that vertices A and B share the same x-coordinate, and vertices B and C share the same y-coordinate.

Equating the y-coordinates of vertices B and C, we have:

4a - 5 = 2a + 6

2a = 11

a = 5.5

Substituting 5.5 for a, we see that the ordered pair for vertex B is (0, 4(5.5) - 5), or (0, 17). Thus, leg AB is the vertical distance from (0,0) to (0,17), or 17.
Similarly, we see that the ordered pair for vertex C is (12, 17) and leg BC is is the horizontal distance from (0,17) to (12, 17) = 12.

Thus, the area of the triangle is:

area = base x height x 1/2

area = 17 x 12 x 1/2 = 17 x 6 = 102

Answer: A
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle  [#permalink]

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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle   [#permalink] 09 Mar 2019, 01:14
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